Problem: Integrate. $ \int 2\sec(x)\tan(x)\,dx $ Choose 1 answer: Choose 1 answer: (Choice A) A $2\sec x + C$ (Choice B) B $-2\sec x + C$ (Choice C) C $-2\sec^3(x)-2\sec(x)\tan^2(x) + C$ (Choice D) D $2\sec^3(x)+2\sec(x)\tan^2(x) + C$
Solution: We need a function whose derivative is $2\sec(x)\tan(x)$. We know that the derivative of $\sec(x)$ is $\sec(x)\tan(x)$, so let's start there: $\dfrac{d}{dx} \sec(x) = \sec(x)\tan(x)$ Now let's multiply by $2$ : $\dfrac{d}{dx} \left[ 2\sec(x) \right]= 2\dfrac{d}{dx} \sec(x) =2\sec(x)\tan(x)$ Because finding the integral is the opposite of taking the derivative, this means that: $ \int 2\sec(x)\tan(x)\,dx =2\sec(x)\, + C$ The answer: $2 \sec(x)\, + C$